Edexcel M4 Exam Discussion Thread [June 5th] Watch

Gome44
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#121
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(Original post by simonli2575)
The one where you are required to find the speed of B.
If it is then I got that question completely wrong. Think I dropped all the marks but 1. Still ended up with an A* on the paper though because the boundary was so low
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simonli2575
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(Original post by Gome44)
If it is then I got that question completely wrong. Think I dropped all the marks but 1. Still ended up with an A* on the paper though because the boundary was so low
Ah no, that one was the easy one. I was talking about the one in which boat B is moving due east and some time later, B is 2sqrt(3) km due south of A.
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Gome44
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(Original post by simonli2575)
Ah no, that one was the easy one. I was talking about the one in which boat B is moving due east and some time later, B is 2sqrt(3) km due south of A.
Oh, I found the difficulty the other way around. I think on the one I sent a photo of I thought P was moving so got that completely wrong. For the one you're talking about though, the angles were like 150 and 210 so it was screaming for I and j vectors
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ThatPerson
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(Original post by Vesniep)
Me too(M3 90,Fp2 100). How did you do in FP2, M3? M4 is not the best unit to rely on (you can aim for A* but not be sure)
If you have M2 and/or FP3 it is better to think of getting a good mark in these units,in my opinion( they are safer)
I'm fairly sure I got 95+ in FP2, and somewhere between 80-90 in M3 depending on the boundaries, so if I take the minimums, to guarantee an A* tomorrow I would need 95 UMS.
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Gome44
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Hoping for a paper like last years (mostly easy but some tricky parts so the grade boundaries fall)
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ThatPerson
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What is the reason for why the component of the velocity that is perpendicular to the common remains unchanged? Also, why does the fact that all collisions in M4 are smooth mean that the mutual reaction acts along the common normal at the point of impact?

If we were solving a problem involving collisions of rough elastic bodies, what would change?
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Gawain
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(Original post by ThatPerson)
What is the reason for why the component of the velocity that is perpendicular to the common remains unchanged? Also, why does the fact that all collisions in M4 are smooth mean that the mutual reaction acts along the common normal at the point of impact?

If we were solving a problem involving collisions of rough elastic bodies, what would change?
When two objects collide, they exert a force upon each other perpendicular to their plane of contact (along the normal of the plane).
This means an impulse is only felt parallel to the line of centres and therefore the velocities perpendicular to this line are unchanged (impulse is 0).

If the objects are completely smooth then there is no force perpendicular to the normal. If they were rough there would be a component frictional force perpendicular to the normal and so both components of velocity would change rather than just one.
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ThatPerson
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(Original post by Gawain)
When two objects collide, they exert a force upon each other perpendicular to their plane of contact (along the normal of the plane).
This means an impulse is only felt parallel to the line of centres and therefore the velocities perpendicular to this line are unchanged (impulse is 0).

If the objects are completely smooth then there is no force perpendicular to the normal. If they were rough there would be a component frictional force perpendicular to the normal and so both components of velocity would change rather than just one.
Thanks for the explanation. So if I have the following question:

"A small smooth spherical ball of mass m falls vertically at strikes a fixed smooth inclined plane with speed u.

(a) Explain why the component of the velocity of the ball parallel to the plane is not affected by the impact"

The answer would be something like "When the collision occurs the ball and plane exert a force on each other that is perpendicular to the plane, hence there is no impulse parallel to the plane and so the parallel velocity is unchanged"?
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TeeEm
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quick question on the relative motion (M4 edexcel)
only a few minutes old ..

I will post the solution real soon
Attached files
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simonli2575
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#130
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(Original post by TeeEm)
quick question on the relative motion (M4 edexcel)
only a few minutes old ..

I will post the solution real soon
How about a question using relative velocities to find the minimum separation given A is trying go get away from B?
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TeeEm
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(Original post by simonli2575)
How about a question using relative velocities to find the minimum separation given A is trying go get away from B?
will try but I do not think it will fit with my to do list before the M4 exam.
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TeeEm
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(Original post by TeeEm)
quick question on the relative motion (M4 edexcel)
only a few minutes old ..

I will post the solution real soon
Question with solution from post 129
Attached files
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jjpneed1
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Can someone please explain question 3 June 2014

Why is 1/2 (x^2 + y^2) = x^2 + y^2/9 ??? If the speed is half then surely  1/2 \sqrt{x^2 + y^2} = \sqrt{x^2 + y^2/9} so 1/4(x^2 + y^2) = x^2 + y^2/9

Am I being really dumb or wtf??
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CowsWhoStare
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Can someone explain to me how to find the least value of speed for something to intercept something else? like in June 2013 question 5


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Gawain
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(Original post by ThatPerson)
Thanks for the explanation. So if I have the following question:

"A small smooth spherical ball of mass m falls vertically at strikes a fixed smooth inclined plane with speed u.

(a) Explain why the component of the velocity of the ball parallel to the plane is not affected by the impact"

The answer would be something like "When the collision occurs the ball and plane exert a force on each other that is perpendicular to the plane, hence there is no impulse parallel to the plane and so the parallel velocity is unchanged"?
Yeah that looks good.


(Original post by CowsWhoStare)
Can someone explain to me how to find the least value of speed for something to intercept something else? like in June 2013 question 5


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To find the minimum velocity for interception, the relative velocity (which is in the direction of the initial distance) is perpendicular to the 'interceptor's' velocity.

For questions relating to crossing a river, the quickest way is to swim straight across (perpendicular to the bank).

(Original post by TeeEm)
Question with solution from post 129
This is just a stylistic question but do you think it is better to separate your diagrams or combine them like this:
Name:  ImageUploadedByStudent Room1433441016.865979.jpg
Views: 94
Size:  154.0 KB
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Gawain
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(Original post by jjpneed1)
Can someone please explain question 3 June 2014

Why is 1/2 (x^2 + y^2) = x^2 + y^2/9 ??? If the speed is half then surely  1/2 \sqrt{x^2 + y^2} = \sqrt{x^2 + y^2/9} so 1/4(x^2 + y^2) = x^2 + y^2/9

Am I being really dumb or wtf??
No that's right,
If you follow on from yours the ratio you get is y/x=(27/5)^0.5 which is what the final line has.

The ratio isn't 3/(7^0.5).
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TeeEm
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(Original post by Gawain)
Yeah that looks good.




To find the minimum velocity for interception, the relative velocity (which is in the direction of the initial distance) is perpendicular to the 'interceptor's' velocity.

For questions relating to crossing a river, the quickest way is to swim straight across (perpendicular to the bank).



This is just a stylistic question but do you think it is better to separate your diagrams or combine then like this:
Name:  ImageUploadedByStudent Room1433441016.865979.jpg
Views: 94
Size:  154.0 KB
everybody has their style and this particular type of question has way too many styles ...
Apologies to any student which finds my style difficult to follow, however feel free to attach an alternative solution if it is going to benefit some students
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Gawain
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(Original post by TeeEm)
everybody has their style and this particular type of question has way too many styles ...
Apologies to any student which finds my style difficult to follow, however feel free to attach an alternative solution if it is going to benefit some students
I don't find your style confusing, to the contrary I think it's clearer. It's just mark schemes combine there diagrams so I thought that's what is expected.
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TeeEm
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(Original post by Gawain)
I don't find your style confusing, to the contrary I think it's clearer. It's just mark schemes combine there diagrams so I thought that's what is expected.
I am not ashamed to admit I cannot fully follow diagrams in marking schemes (or indeed some of the diagrams in the examples in the new book of EDEXCEL).
The way I present it is the way I understood the topic as a learner a very long time ago. (Fashions do change)
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jjpneed1
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(Original post by Gawain)
No that's right,
If you follow on from yours the ratio you get is y/x=(27/5)^0.5 which is what the final line has.

The ratio isn't 3/(7^0.5).
Using dot product on the velocity before and after gives the wrong answer for some reason though, but yeah the ratio is still right.

Also another problem with Q5 on the same paper

In the conservation of momentum calculation I used x and y going in the opposite direction to that shown in the diagram in the mark scheme, so

-u = x + 3y and x - y = u/3

but this gives x=0, y=-u/3 which is wrong... why would it matter which why I label the velocities...? Somehow I went through the other mechanics modules without encountering this problem

EDIT: nvm sign error using restitution I believe
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